CPM Homework Help : APCALC Problem 3-64

Using the definition of the derivative as a limit, show that the derivative of $f(x) = \frac { 1 } { x ^ { 2 } }$ is $f^\prime (x) = -\frac { 2 } { x ^ { 3 } }$ . That is, show algebraically that the following limit statement is true:

$\lim\limits _ { h \rightarrow 0 } \frac { \frac { 1 } { ( x + h ) ^ { 2 } } - \frac { 1 } { x ^ { 2 } } } { h } = - \frac { 2 } { x ^ { 3 } }$

Hint:

The Derivative

The slope of a line tangent to at any point $x$ is called the derivative of $f$ at $x$ . It is found by taking a limit of the slope of a secant line as $h \rightarrow 0$ . The standard form of this type of limit is:

$f^\prime (x) = \lim\limits _ { h \rightarrow 0 } \frac { f ( x + h ) - f ( x ) } { h }$

If the slope (or instantaneous rate of change) at a particular $x$ -value is desired, such as at $x = a$ , then the notation used is $f^\prime (a)$ . This slope can be found by replacing $x$ with $a$ .

$f^\prime (a) = \lim\limits _ { h \rightarrow 0 } \frac { f ( a + h ) - f ( a ) } { h }$

First quadrant downward parabola, labeled, f of x, 1 tick mark on x axis, labeled, a, with highlighted point on curve, corresponding to tick mark, solid increasing line, labeled slope = f prime of a, tangent to the curve at the corresponding point, & multiple dashed lines, all passing through the corresponding point, & various other points on the curve.

Hint:

This is one form of the 'definition of the derivative' (informally known as Hana's Method).

Hint:

In order to evaluate this limit, we need to find an Algebraic way to cancel out the $h$ in the denominator.

Step 1:

Find a common denominator in the numerator, expand and combine like terms:

$\lim\limits_{h\rightarrow 0}\frac{\frac{x^{2}-(x+h)^{2}}{(x^{2})(x+h)^{2}}}{h}=\lim\limits_{h\rightarrow 0}\frac{x^{2}-(x^{2}+2xh+h^{2})}{h(x^{2})(x+h)^{2}}=\lim\limits_{h\rightarrow 0}\frac{-2xh-h^{2}}{h(x^{2})(x+h)^{2}}$

Step 2

Factor the numerator, then cancel out the $h$ :

$\lim\limits_{h\rightarrow 0}\frac{h(-2x-h)}{h(x^{2})(x+h)^{2}}=\lim\limits_{h\rightarrow 0}\frac{(-2x-h)}{(x^{2})(x+h)^{2}}$

Answer:

Since there is no longer and $h$ in the denominator, you can evaluate the limit as $h \rightarrow 0$ :

$=\lim\limits_{h\rightarrow 0}\frac{(-2x-(0))}{(x^{2})(x+(0))^{2}}=\lim\limits_{h\rightarrow 0}\frac{(-2x)}{(x^{2})(x)^{2}}=-\frac{2}{x^{3}}$